If we are allowed to use basic properties of logarithms, the problem is straightforward. Let $a=\lfloor \log x\rfloor$. Then $\log x=a+e$, where $0\le e\lt 1$.

The inequality on the right is obvious. For the inequality on the left, note that
$$x=2^{\log x}=2^a2^e.$$
But $2^e\lt 2$. It follows that $2^a\gt \frac{1}{2}x$. If we do not wish to use the word "obvious" for the inequality on the right, use $1\le 2^e\lt 2$.

I suspect what your professor expects you to do is expand that middle term:

$ 2^{log(x)} = x ^{log(2)}$

Then you just have to show that

$ \frac{x}{2} < x^{log(2)} \leq x $

The easiest way to get that done is to actually show it's true for a base case (like x = 1), then show that, for $ x \geq 1 $, the inequality for x implies the inequality for x +1 (i.e. proceed by induction).

Actually, it may be easier to show that it's true for x = 1 and x = 2, and then proceed by induction assuming $ x \geq 2 $.

Yes, I started with the base case, x=1, but I can't get it for x+1
–
eoutiOct 8 '12 at 20:41

Then I suggest you use a second base case, $ x = 2 $. It should be easier to show that $ \frac{x}{2} < x^{log(2)} \leq x \Rightarrow \frac{x+1}{2} < (x+1)^{log(2)} \leq x +1 $ if you assume that $ x \geq 2 $ (which you can, by using a second base case).
–
Mathew CalkinsOct 8 '12 at 20:54